Question: Is ${964523}$ divisible by $4$ ?
A number is divisible by $4$ if the last two digits are divisible by $4$ . [ Why? We can rewrite the number as a multiple of $100$ plus the last two digits: $ \gray{9645} {23} = \gray{9645} \gray{00} + {23} $ Because $964500$ is a multiple of $100$ , it is also a multiple of $4$ So as long as the value of the last two digits, ${23}$ , is divisible by $4$ , the original number must also be divisible by $4$ Is the value of the last two digits, $23$ , divisible by $4$ No, $23$ is not divisible by $4$, so $964523$ is also not divisible by $4$.